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Let $X,Y$ be Hilbert spaces. I just proved that, given $A: X\rightarrow Y$ a bounded operator, one has:$$\ker(A^*)_\perp=\ker(A^t)^\perp$$

Here $U_\perp$ means pre-annihilator, $V^\perp$ means orthogonal complement, $A^*: Y^*\rightarrow X^*$ is the dual of $A$, $A^t: Y\rightarrow X$ the Hilbert space adjoint of $A$.

My proof distinguishes the two inclusions, and uses the definition of $(\cdot)^t$ and the Riesz representation theorem multiple times to compare the notions of pre-annihilator and orthogonal complement. However it is quite long winded and I feel there is a sleek way of seeing this result, which I couldn't find.

Do you have any idea?

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    $\begingroup$ Most likely the proof you have is the best one. $\endgroup$ Commented Feb 25, 2021 at 11:44

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Let $\phi\leftrightarrow y_\phi$ be the Riesz correspondence on $Y^*$.

Then for all $\phi\in Y^*$, by definition of the Hilbert space adjoint, \begin{align}A^*\phi=0&\iff (\forall x\in X, \phi(Ax)=(A^*\phi) x=0)\\ &\iff (\forall x\in X, \langle y_\phi,Ax\rangle=0)\\&\iff A^ty_\phi=0.\end{align}

Hence, since $\phi\in Y^*$ correspond to $y\in Y$ by the Riesz representation, \begin{align} x\in(\ker A^*)_\perp&\iff \forall\phi\in Y^*,(A^*\phi=0\implies\phi(x)=0)\\ &\iff \forall \phi\in Y^*, (A^ty_\phi=0\implies\langle y_\phi,x\rangle=0)\\ &\iff \forall y\in Y, (A^ty=0\implies \langle y,x\rangle=0)\\ &\iff \forall y\in Y, (y\in \ker A^t\implies y\perp x)\\ &\iff x\in(\ker A^t)^\perp\end{align}

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